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**SOgLak** · 09-12-2007, 07:08 PM

That's a funny way of saying it.
Obviously the solution doesn't have to be polynomial.

Yes it does, that's guaranteed by how I got q and r and p.

PLUS, the q[x+1] in the notation is unneeded as far as I can tell, since you might as well assume it's q[x].

It's needed when you know the origin of those three polynomials.

(For context, I'm performing Gosper's algorithm to find the closed form of a hypergeometric sum, if it exists. This is just one step.)

09-12-2007, 07:08 PM	#6
SOgLak Join Date Oct 2005 Posts 377 Senior Member	That's a funny way of saying it. Obviously the solution doesn't have to be polynomial. Yes it does, that's guaranteed by how I got q and r and p. PLUS, the q[x+1] in the notation is unneeded as far as I can tell, since you might as well assume it's q[x]. It's needed when you know the origin of those three polynomials. (For context, I'm performing Gosper's algorithm to find the closed form of a hypergeometric sum, if it exists. This is just one step.)
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